Last visit was: 08 Sep 2026, 14:28

It is currently 08 Sep 2026, 14:28

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Sep 09
Masters' Spotlight Fair
07:00 AM PDT
-
10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway
Sep 09
How Aman Bounced Back from 3 MBA Rejections to Get Into Cambridge
10:00 AM PDT
-
11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application...
Sep 04
BIG NEWS! We’ve been nominated, and we’d love your vote!
10:00 AM EDT
-
11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.

Back to Forum

Create Topic

GMAT Club World Cup 2026 (Day 3): Three buses leave Riverton and trave

1 2 3 4 5 6 7 8

omnisin

Joined: 04 Aug 2025

Last visit: 13 Aug 2026

Posts: 14

Own Kudos:

[1]

Given Kudos: 5

Posts: 14

Kudos: 10

[1]

08 Jul 2026, 14:52

Kudos

Bookmarks

Three buses leave Riverton and travel along the same route toward Lakeview. Bus 1 leaves first and travels at a constant speed of 50 km/h. Bus 2 leaves Riverton 6 hours after Bus 1 and travels at a constant speed of 80 km/h. Bus 3 also leaves Riverton after Bus 1 and travels at a constant speed of 100 km/h. If Bus 2 and Bus 3 catch up to Bus 1 at the same time, how many hours after Bus 1 leaves does Bus 3 leave?

A. 2
B. 6
C. 8
D. 10
E. 16

We know that after 6 hours, but one will be 300 km away (50 km/h x 6 hours), and bus 2 is just leaving. Bus 2 travels at 80 km/h, so it will gain 30 km/h on bus one since 80-50 km/h =30. This means it will take take 10 hours for bus 2 to catch up to bus 1, having traveled a total of 800 miles. 50 km/h x16 hrs= 80 km/h x10 hrs.

To travel the same 800 miles, it will take a total 8 hrs since 800 km/100 km per h is 8. Since bus 1 takes a total of 16 hrs, this means bus 3 leaves 8 hrs afterwards. Option c.

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

c404

Joined: 18 Jan 2026

Last visit: 08 Sep 2026

Posts: 10

Own Kudos:

[1]

Given Kudos: 84

Posts: 10

Kudos: 10

[1]

08 Jul 2026, 14:57

Kudos

Bookmarks

B1 travels 300 km before B2 start
B2 is 30km/h faster so it takes 10 hours to cover that B1 head start. Then total time is 10+6 = 16 hours

while total B1 distance = 50 x 16 = 800km
which B3 can cover in 8 hours, so it can give B1 a 8 hour head start

RS2302

Joined: 08 Nov 2023

Last visit: 16 Aug 2026

Posts: 16

Own Kudos:

[1]

Given Kudos: 5

Location: India

Posts: 16

Kudos: 11

[1]

08 Jul 2026, 15:28

Kudos

Bookmarks

For convenience purpose, I will call Buses as B1, B2, B3 in the order they are given.

B1 leaves and travels for 6 hours before B2 starts. Hence, by the time B2 leaves, B1 has already travelled 6x50=300.

Now, B2 has to cover 300km in order to catch up with B1.

Using relative speed formula we can find out how much time B2 took before it could catch up to B1.
————-> 300/80-50 = 10 hours.

At this point B3 has also caught up with B1. Now, B1 must have travelled 300km + (10x50)km=800 km

To travel 800km, time taken by B3 is going to be 800/100=8 hours.
And B1 has travelled 16 hours

That means 16-8=8 . B3 left 8 hours after B1 left.

ga821

Joined: 06 Jun 2021

Last visit: 08 Sep 2026

Posts: 58

Own Kudos:

[1]

Given Kudos: 74

Products:

Posts: 58

Kudos: 33

[1]

08 Jul 2026, 16:07

Kudos

Bookmarks

Common multiples of 50, 80 and 100 is 400, 800, and so on

Lets assume distance when they meet to be :

400 - then Bus 2 would have travelled 5km and Bus 3 would have travelled 4 km
but Bus 1 would have travelled 5+6 = 11 hrs which comes to 50 kph * 11 hr=550 km.. this doesn't tally with 400 therefore this is not the distance when they meet

Likewise lets take 800,
Bus 2 would have travelled 10 hrs, Bus 3 would have travelled 8 hrs
and Bus 1 would have travlled 10+6 = 16 hrs which comes to 50 kph + 16 hrs = 800 km

Since this tallies, all 3 met when they had travelled 800 km
Therefore
Bus 3 hours - Bus 1 hour when they meet = 16-8 = 8 hrs

Hence Bus 3 left 8 hours after Bus 1

Ans - C

FreshStove

Joined: 22 Jun 2026

Last visit: 09 Jul 2026

Posts: 31

Own Kudos:

Given Kudos: 63

Location: India

Schools: ISB

GPA: 9.14

Schools: ISB

Posts: 31

Kudos: 8

08 Jul 2026, 16:30

Kudos

Bookmarks

Using t = d/s2-s1, we first calculated that Bus 1 covered 50*6 = 300km, now d = 300, S1 = 50 and S2 = 80, thus, T = 10 hrs.
Equating 10 = d/100-50 (Bus 3 and Bus 1) speed to arrive at 500km. Thus, Bus 3 leaves 10 hrs after Bus 1.

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

JayBKK

Joined: 02 Jan 2025

Last visit: 08 Sep 2026

Posts: 40

Own Kudos:

Given Kudos: 42

Location: Thailand

Concentration: Entrepreneurship, Economics

Schools: Stanford HBS

GMAT Focus 1: 555 Q82 V77 DI72 Verified score

GPA: 3.37

WE:Engineering (Technology)

Schools: Stanford HBS

GMAT Focus 1: 555 Q82 V77 DI72 Verified score

Posts: 40

Kudos: 29

08 Jul 2026, 17:31

Kudos

Bookmarks

Assume t is the number of hours it takes the 3rd bus to catch up. We know it spent the same amount of time as the 2nd bus did catching up to the 1st bus.

So, we're finding the time when the 2nd bus caught up with the first.

Find the distance of the leading bus: s = 50 × 6 = 300 km.

Find the time it took the second bus to catch up as a relative velocity: t = 300 / (80-50) = 10 hrs.

So, we know that t, the time it takes the third bus to catch up, is 10 hrs plus 6 hrs while the second bus is waiting, for a total of 16 hrs.

Choice E.

scared

Joined: 19 Dec 2022

Last visit: 08 Sep 2026

Posts: 2

Own Kudos:

[1]

Given Kudos: 2

Posts: 2

Kudos: 2

[1]

08 Jul 2026, 17:34

Kudos

Bookmarks

the distance traveled by all 3 should be same.
so
50(6+x)=80x where x is the time bus 2 traveled for
x= 10.
bus 1 traveled for 6+10 = 16 hrs.
50*16 = 100*y where y is the time bus 3 traveled for
y=8
thus the bus 3 started 16-8 = 8 hrs after bus 1.

rockstar09rulez

Joined: 30 Sep 2023

Last visit: 06 Sep 2026

Posts: 43

Own Kudos:

[1]

Given Kudos: 74

Location: United States (CA)

Concentration: Technology, Finance

Products:

Posts: 43

Kudos: 25

[1]

08 Jul 2026, 17:42

Kudos

Bookmarks

B1 - 50kmph, B2 - 80kmph, B3 - 100kmph
Since B2 starts 6 hours later, B1 will have covered 6*50 = 300km in those 6 hours

Now using relative speed to find catchup time for B2
relative speed = B2-B1 = 30kmph, D = 300km, T = 300/30 = 10 hours
So, in 10 hours - B2 will catchup to B1
This means B1 will have travelled 10 more hours after it had already travelled 6 hours, so total 16 hours.

Now B2 will have covered 80*10 = 800km in 10 hours

Given B3 also catches up at the same time - meaning it would also have covered 800km
B3 covered 800km in 800/100 = 8 hours

So, B1 covered that distance in 16 hours, while B3 did it in 8 hours. So, B3 would have started 8 hours after B1 --- ANS (C)

snorlax02

Joined: 22 Dec 2025

Last visit: 30 Jul 2026

Posts: 39

Own Kudos:

[1]

Given Kudos: 4

Posts: 39

Kudos: 24

[1]

08 Jul 2026, 19:09

Kudos

Bookmarks

Assuming Buses B2 & B3 meet after d distance.

d = 50 * t (t - time taken for B1 to reach d distance)
d = 80 * (t-6) (Because B2 leaves after 6 hours)

50t = 80(t-6) => t = 16hrs & d = 800 kms

since B3 is travelling at 100km/hr to travel 800Km it takes 8 hrs. Thus to catch up at the same time it leaves after (16 - 8) i.e 8 hrs

Ans - C

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

Triloki

Joined: 26 May 2026

Last visit: 08 Sep 2026

Posts: 5

Own Kudos:

[1]

Given Kudos: 1

Posts: 5

Kudos: 6

[1]

08 Jul 2026, 19:16

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

Let bus 2/3 catch bus 1 after t hr from start of bus 1 and bus 3 start after x hr from start of 1.
Distance traveled by 1 = 50t
Distance traveled by 2 = 80 (t-6)
Distance traveled by 3 = 100(t-x)
Equating all
T = 16 and x =8
Ans is 8

BongBideshini

Joined: 25 May 2021

Last visit: 08 Sep 2026

Posts: 391

Own Kudos:

244

[1]

Given Kudos: 137

Location: India

Concentration: Entrepreneurship, Marketing

Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27

GPA: 3.8

WE:Engineering (Government)

Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27

Posts: 391

Kudos: 244

[1]

08 Jul 2026, 19:18

Kudos

Bookmarks

Let's say bus 1 took t hrs
50t=(t-6)*80
t=16

Bus 3's speed is twice the speed of Bus 1 while the distance is the same.
So time taken will be 1/2 of the bus-1 's time
Bus-3 took 8hrs, so it left (16-8)=8 hrs after bus-1

Ans C

DishaSaha

Joined: 03 Nov 2024

Last visit: 30 Aug 2026

Posts: 64

Own Kudos:

[1]

Given Kudos: 24

GMAT Focus 1: 555 Q81 V75 DI78 Verified score

Posts: 64

Kudos: 49

[1]

08 Jul 2026, 19:47

Kudos

Bookmarks

Bus	Speed	time	distance travelled
1	50	t(say)	50t
2	80	t-6	80(t-6)
3	100	x(say)	100x

Considering Bus 1 & 2,
Since, they meet at time t since bus 1 leaves,
.:50t=80(t-6),
.:30t=480
.:t=16 hr.

Since Bus3 also met at the same time,

50*16=100*x
.: x=8 hrs

.: Bus 3 left after (16-8)=8 hrs Bus1 left.

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

PJ8

Joined: 08 Jun 2026

Last visit: 30 Jul 2026

Posts: 21

Own Kudos:

[1]

Given Kudos: 6

Posts: 21

Kudos: 13

[1]

08 Jul 2026, 19:48

Kudos

Bookmarks

Since Bus 1,2,3 met at the same time, so finding distance and Time travelled by Bus 1 is necessary to find the delay hours at Bus 3 starts.
So 1st equation would be:
50(T+6) = 80T ; T = 10 Hr ( T is running time Bus 2)
Total distance covered by Bus 1 = 50(10+6) = 800Km
800 Km would be covered by Bus 3 @100Km/h = 8 Hr
Since Bus 1 & 3 met at same time so
time period after which Bus 3 starts after Bus 1 = Time for which Bus 1 ran - Time taken by Bus 3 to cover 800 Km distance = 16-8 = 8.

patnaik1988

Joined: 05 Oct 2024

Last visit: 08 Sep 2026

Posts: 84

Own Kudos:

[1]

Given Kudos: 68

Posts: 84

Kudos: 19

[1]

08 Jul 2026, 19:51

Kudos

Bookmarks

Ans Choice C : Bus2 when starts Bus1 would have travelled 50x6=300 km. To cover this distance Bus2 would take 300/(80-50)=10hrs.
So total time when Bus1 & 2 meet = 6+10 = 16hrs (from Start).....(1)
Let Bus3 start after x hrs from Bus1;
Bus1 would travel = 50x kms
in order to cover this distance Bus3 would take 50x/(100-50) = x hrs
Total time from start = x + x = 2x hrs.....(2)
Equating 1&2 gives x=8hrs. (Ans Choice C)

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

harshitaa45

Joined: 20 Feb 2021

Last visit: 08 Sep 2026

Posts: 49

Own Kudos:

[1]

Given Kudos: 8

Products:

Posts: 49

Kudos: 30

[1]

08 Jul 2026, 20:33

Kudos

Bookmarks

BUS 1 leaves first at a speed of 50 kmph.
Bus 2 starts 6 hrs later at speed 80kmph: in this time the bus 1 would have covered :300 km
SO let h hrs is what bus 2 takes to catch Bus1:
50h+300=80h => h=10.
So the time it takes for bus 2 to meet bus 1 is 16 hrs.
It is also given that bus 2 and bus 3 catches bus 1 simultaneously.
In 16hrs the distance traveled by Bus 1 is 16*50=800 kms

time it takes for Bus 3 to travel 800 kms is 800/100=8 hrs.
So Bus 3 left 8 hrs after Bus 1 departure

Aarushi100

Joined: 17 Jul 2024

Last visit: 07 Sep 2026

Posts: 70

Own Kudos:

[1]

Given Kudos: 98

Posts: 70

Kudos: 57

[1]

08 Jul 2026, 20:41

Kudos

Bookmarks

Now, let's assume Bus1 meets Bus2 at time t.
Both buses travel the same distance by then:
50*t = 80(t-6)
=> t= 16.
Bus2 catches up 16hours later.

Now Bus3 also catches up with Bus1, 16hours after Bus1 leaves.
So,
Let x be the number of hours later Bus3 starts.
100(16-x) =50*16
=> x=8
ANSWER C

metrogloomin

Joined: 16 Mar 2026

Last visit: 06 Sep 2026

Posts: 33

Own Kudos:

[1]

Given Kudos: 84

Location: United States (NY)

Schools: CBS '27 Booth '27 Wharton '27 Stern '27 Sloan '27 Stanford '27 HBS '27

GPA: 7.05

Products:

Schools: CBS '27 Booth '27 Wharton '27 Stern '27 Sloan '27 Stanford '27 HBS '27

Posts: 33

Kudos: 31

[1]

08 Jul 2026, 20:42

Kudos

Bookmarks

Speeds of Bus 1, 2, 3 respectively are 50 km/h, 80 km/h, 100 km/h.
We know all three buses are meeting at the same distance from the start, at the same time irrespective of start time.

Bus 1 has travelled the entire time at 50 km/h.
Bus 2 has a lag of 6 hours, and let us assume it travelled for x hours making the distance covered 80x km.
By the time Bus 2 covered 80x km, Bus 1 had been travelling for (x+6) hours. Hence, distance covered by Bus 1 in the same duration=50(x+6)=50x+300 km.
Since Bus 2 and Bus 1 have caught up with each other, both distances are equal. Therefore, 80x=50x+300, making x=10 hours.

We know the total distance where all three buses have met is 80x km and x=10 hours, thus 80x=800 km.
Bus 3 would take 8 hours to cover 800 km, and hence it has travelled for 8 hours.
Bus 1 has travelled for (x+6) hours, i.e. 16 hours.
Thus, Bus 3 left after (16-8)=8 hours (C)

Sumimasen

Joined: 21 Jan 2024

Last visit: 27 Jul 2026

Posts: 48

Own Kudos:

[1]

Given Kudos: 13

Posts: 48

Kudos: 43

[1]

08 Jul 2026, 21:12

Kudos

Bookmarks

In 6 hours, bus1 covers 300 km. This distance will be covered by bus2 in 10 hours.
300/(80-50)= 300/30 = 10 hours.

In 10+6 hours bus 1 has travelled total distance of = 50x16 = 800km
These 800km will be covered by bus3 in 800/100 = 8 hours. Bus 3 travels for 8 hours and catch up to bus1 after 16 hours from bus 1 departure. This means bus 3 has left after 8 hours from bus 1 departure.
16-8 = 8hours

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

Vijendra.D

Joined: 12 Jun 2024

Last visit: 13 Aug 2026

Posts: 27

Own Kudos:

[1]

Given Kudos: 148

GMAT Focus 1: 575 Q81 V81 DI74 Verified score

GMAT Focus 2: 615 Q85 V78 DI78 Verified score

Posts: 27

Kudos: 15

[1]

08 Jul 2026, 21:22

Kudos

Bookmarks

Considering buses 1 and 2

Bus 2 starts after 6 hrs..
At that time Bus 1 travelled a total of 300 kms.

Now Consider time required for Bus2 to catch Bus1

Relative Speed== 80 - 50 == 30kmph
The distance between them is 300km
Therefore bus2 needs additional 10 hrs to catch bus 1.

Total distance covered by bus 1 after (6hrs + 10hrs) = 16hrs is 800km.

Now bus3 needs only 8 hrs to cover 800km.

Total time bus 3 needs is 8 hrs to catch up

So the difference of time between bus1 and bus 3 is 16-8 == 8 hrs

GMATNextStep

Joined: 01 Aug 2023

Last visit: 08 Sep 2026

Posts: 40

Own Kudos:

[1]

Given Kudos: 1

GMAT Focus 1: 715 Q85 V88 DI84 (Online) Verified score

Posts: 40

Kudos: 41

[1]

08 Jul 2026, 22:01

Kudos

Bookmarks

Bunuel

This question was provided by GMAT Club
for the GMAT World Cup Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.

My Answer is C.

Time catch up = Delta Distance/ Delta Rate

Time for Bus 2 to catch up to Bus 1 = (50x6)/(80-50) = 10 hours
=> total time Bus 1 traveled is 10+6 = 16 hours => Total distance Bus 1 traveled is 50x16 = 800 km

Since we know Bus 3 also catches up to Bus 1 at the same time as Bus 2, it means that the total distance Bus 3 travels will be equal to that of Bus 1.
=> total time Bus 3 traveled is 800/100 = 8 hours. Since by the time they meet up, Bus 1 will have traveled for 16 hours while Bus 3 traveled only for 8 hours, meaning Bus 3 started 8 hours after